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For the arrangement shown in Fig. 2. find (a) The equivalent capacitance of the circuit, (b) The voltage across QR, (c) the charge on each capacitor

Physics

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4. Diagram of Leclanché Cell and its Parts

A Leclanché cell is a primary cell that converts chemical energy into electrical energy. It consists of:

  • A carbon rod (positive electrode) surrounded by a mixture of manganese dioxide (MnO2MnO_2) and carbon powder (depolarizer) packed in a porous pot.
  • A zinc container (negative electrode).
  • An ammonium chloride (NH4ClNH_4Cl) solution (electrolyte) that fills the space between the zinc container and the porous pot.

5. Circuit Analysis (Fig 1)

a) The total resistance

  1. Calculate the equivalent resistance of the series combination of 8Ω, 7Ω, and 5Ω resistors.

    Rseries=8 Ω+7 Ω+5 Ω=20 ΩR_{series} = 8\, \Omega + 7\, \Omega + 5\, \Omega = 20\, \Omega

  2. Calculate the equivalent resistance of RseriesR_{series} (20Ω) in parallel with the 18Ω resistor.

    1Rparallel=120 Ω+118 Ω\frac{1}{R_{parallel}} = \frac{1}{20\, \Omega} + \frac{1}{18\, \Omega} 1Rparallel=9180 Ω+10180 Ω=19180 Ω\frac{1}{R_{parallel}} = \frac{9}{180\, \Omega} + \frac{10}{180\, \Omega} = \frac{19}{180\, \Omega} Rparallel=18019 Ω≈9.4737 ΩR_{parallel} = \frac{180}{19}\, \Omega \approx 9.4737\, \Omega

  3. Calculate the total resistance by adding RparallelR_{parallel} and the 30Ω resistor (they are in series).

    Rtotal=Rparallel+30 ΩR_{total} = R_{parallel} + 30\, \Omega Rtotal=18019 Ω+30 Ω=18019 Ω+57019 Ω=75019 ΩR_{total} = \frac{180}{19}\, \Omega + 30\, \Omega = \frac{180}{19}\, \Omega + \frac{570}{19}\, \Omega = \frac{750}{19}\, \Omega Rtotal≈39.47 ΩR_{total} \approx 39.47\, \Omega

    The total resistance is 39.47 Ω\boxed{39.47\, \Omega}.

    b) The voltage drop across the 7Ω resistor

  4. Calculate the total current (ItotalI_{total}) flowing from the 12V battery using Ohm's Law.

    Itotal=VsourceRtotalI_{total} = \frac{V_{source}}{R_{total}} Itotal=12 V75019 Ω=12×19750 A=228750 A=38125 AI_{total} = \frac{12\, \text{V}}{\frac{750}{19}\, \Omega} = \frac{12 \times 19}{750}\, \text{A} = \frac{228}{750}\, \text{A} = \frac{38}{125}\, \text{A} Itotal=0.304 AI_{total} = 0.304\, \text{A}

  5. Calculate the voltage across the parallel combination (VparallelV_{parallel}).

    Vparallel=Itotal×RparallelV_{parallel} = I_{total} \times R_{parallel} Vparallel=38125 A×18019 Ω=2×180125 V=360125 V=7225 VV_{parallel} = \frac{38}{125}\, \text{A} \times \frac{180}{19}\, \Omega = \frac{2 \times 180}{125}\, \text{V} = \frac{360}{125}\, \text{V} = \frac{72}{25}\, \text{V} Vparallel=2.88 VV_{parallel} = 2.88\, \text{V}

  6. Calculate the current through the series branch containing the 7Ω resistor (Iseries_branchI_{series\_branch}).

    Iseries_branch=VparallelRseriesI_{series\_branch} = \frac{V_{parallel}}{R_{series}} Iseries_branch=7225 V20 Ω=7225×20 A=72500 A=18125 AI_{series\_branch} = \frac{\frac{72}{25}\, \text{V}}{20\, \Omega} = \frac{72}{25 \times 20}\, \text{A} = \frac{72}{500}\, \text{A} = \frac{18}{125}\, \text{A} Iseries_branch=0.144 AI_{series\_branch} = 0.144\, \text{A}

  7. Calculate the voltage drop across the 7Ω resistor (V7ΩV_{7\Omega}).

    V7Ω=Iseries_branch×7 ΩV_{7\Omega} = I_{series\_branch} \times 7\, \Omega V7Ω=18125 A×7 Ω=126125 VV_{7\Omega} = \frac{18}{125}\, \text{A} \times 7\, \Omega = \frac{126}{125}\, \text{V} V7Ω=1.008 VV_{7\Omega} = 1.008\, \text{V}

    The voltage drop across the 7Ω resistor is 1.008 V\boxed{1.008\, \text{V}}.

    c) The current through the 30Ω resistor

    The 30Ω resistor is in series with the parallel combination, so the current through it is the total current from the battery.

    I30Ω=Itotal=0.304 AI_{30\Omega} = I_{total} = 0.304\, \text{A}

    The current through the 30Ω resistor is 0.304 A\boxed{0.304\, \text{A}}.

    d) The power developed in the 8Ω resistor

    The current through the 8Ω resistor is Iseries_branchI_{series\_branch}.

    P8Ω=Iseries_branch2×R8ΩP_{8\Omega} = I_{series\_branch}^2 \times R_{8\Omega} P8Ω=(18125 A)2×8 ΩP_{8\Omega} = \left(\frac{18}{125}\, \text{A}\right)^2 \times 8\, \Omega P8Ω=32415625 A2×8 Ω=259215625 WP_{8\Omega} = \frac{324}{15625}\, \text{A}^2 \times 8\, \Omega = \frac{2592}{15625}\, \text{W} P8Ω≈0.166 WP_{8\Omega} \approx 0.166\, \text{W}

    The power developed in the 8Ω resistor is 0.166 W\boxed{0.166\, \text{W}}.

    e) The power developed by the circuit

    The total power developed by the circuit is the product of the total voltage and the total current.

    Ptotal=Vsource×ItotalP_{total} = V_{source} \times I_{total} Ptotal=12 V×38125 A=456125 WP_{total} = 12\, \text{V} \times \frac{38}{125}\, \text{A} = \frac{456}{125}\, \text{W} Ptotal=3.648 WP_{total} = 3.648\, \text{W}

    The power developed by the circuit is 3.648 W\boxed{3.648\, \text{W}}.

    6. Capacitor Arrangement (Fig 2)

    a) The equivalent capacitance of the circuit

  8. Calculate the equivalent capacitance of the parallel combination of 2µF and 3µF capacitors.

    Cparallel=C1+C2C_{parallel} = C_1 + C_2 Cparallel=2 μF+3 μF=5 μFC_{parallel} = 2\, \mu\text{F} + 3\, \mu\text{F} = 5\, \mu\text{F}

  9. Calculate the equivalent capacitance of CparallelC_{parallel} (5µF) in series with the 15µF capacitor.

    1Ceq=1Cparallel+1C3\frac{1}{C_{eq}} = \frac{1}{C_{parallel}} + \frac{1}{C_3} 1Ceq=15 μF+115 μF\frac{1}{C_{eq}} = \frac{1}{5\, \mu\text{F}} + \frac{1}{15\, \mu\text{F}} 1Ceq=315 μF+115 μF=415 μF\frac{1}{C_{eq}} = \frac{3}{15\, \mu\text{F}} + \frac{1}{15\, \mu\text{F}} = \frac{4}{15\, \mu\text{F}} Ceq=154 μF=3.75 μFC_{eq} = \frac{15}{4}\, \mu\text{F} = 3.75\, \mu\text{F}

    The equivalent capacitance of the circuit is 3.75 μF\boxed{3.75\, \mu\text{F}}.

    b) The voltage across QR

    The voltage across QR is the voltage across the 15µF capacitor.

  10. Calculate the total charge (QtotalQ_{total}) stored in the equivalent capacitance.

    Qtotal=Ceq×VsourceQ_{total} = C_{eq} \times V_{source} Qtotal=3.75 μF×240 VQ_{total} = 3.75\, \mu\text{F} \times 240\, \text{V} Qtotal=900 μCQ_{total} = 900\, \mu\text{C}

  11. For capacitors in series, the charge on each equivalent capacitor is the same.

    So, the charge on the 15µF capacitor is QtotalQ_{total}.

    Q15μF=900 μCQ_{15\mu F} = 900\, \mu\text{C}

  12. Calculate the voltage across the 15µF capacitor (VQRV_{QR}).

    VQR=Q15μFC15μFV_{QR} = \frac{Q_{15\mu F}}{C_{15\mu F}} VQR=900 μC15 μF=60 VV_{QR} = \frac{900\, \mu\text{C}}{15\, \mu\text{F}} = 60\, \text{V}

    The voltage across QR is 60 V\boxed{60\, \text{V}}.

    c) The charge on each capacitor

  13. Charge on the 15µF capacitor (Q15μFQ_{15\mu F}).

    As calculated in part (b), Q15μF=900 μCQ_{15\mu F} = 900\, \mu\text{C}.

  14. Voltage across the parallel combination (VparallelV_{parallel}).

    The total voltage is 240V. The voltage across the 15µF capacitor is 60V.

    Vparallel=Vsource−V15μFV_{parallel} = V_{source} - V_{15\mu F} Vparallel=240 V−60 V=180 VV_{parallel} = 240\, \text{V} - 60\, \text{V} = 180\, \text{V}

  15. Charge on the 2µF capacitor (Q2μFQ_{2\mu F}).

    Q2μF=C2μF×VparallelQ_{2\mu F} = C_{2\mu F} \times V_{parallel} Q2μF=2 μF×180 V=360 μCQ_{2\mu F} = 2\, \mu\text{F} \times 180\, \text{V} = 360\, \mu\text{C}

  16. Charge on the 3µF capacitor (Q3μFQ_{3\mu F}).

    Q3μF=C3μF×VparallelQ_{3\mu F} = C_{3\mu F} \times V_{parallel} Q3μF=3 μF×180 V=540 μCQ_{3\mu F} = 3\, \mu\text{F} \times 180\, \text{V} = 540\, \mu\text{C}

    The charge on each capacitor is:

    • 15µF capacitor: 900 μC\boxed{900\, \mu\text{C}}
    • 2µF capacitor: 360 μC\boxed{360\, \mu\text{C}}
    • 3µF capacitor: 540 μC\boxed{540\, \mu\text{C}}

    2 zimekamilika. Zimebaki 3 leo — tuma inayofuata.

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